Answer:
Yes that's right. Not really sure what your looking for as far as a reason goes. You performed the correct operations and got the right answer.
The change in the player's internal energy is -491.6 kJ. The number of nutritional calories is -117.44 kCal
For this process to take place, some of the basketball player's perspiration must escape from the skin. This is because sweating relies on a physical phenomenon known as the heat of vaporization.
The heat of vaporization refers to the amount of heat required to convert 1g of a liquid into a vapor without causing the liquid's temperature to increase.
From the given information,
- the work done on the basketball is dW = 2.43 × 10⁵ J
The amount of heat loss is represented by dQ.
where;
∴
Using the first law of thermodynamics:b
dU = dQ - dW
dU = -mL - dW
dU = -(0.110 kg × 2.26 × 10⁶ J/kg - 2.43 × 10⁵ J)
dU = -491.6 × 10³ J
dU = -491.6 kJ
The number of nutritional calories the player has converted to work and heat can be determined by using the relation:

dU = -117.44 kcal
Learn more about first law of thermodynamics here:
brainly.com/question/3808473?referrer=searchResults
Answer:
22cm
Explanation:
focal length = 11cm
radius of curvature,r = 2f
r= 2 x 11
r=22cm
Answer:
what is heat and transfer
Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K